Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000000 + 7;
long long powe(long long x, long long y) {
  x = x % mod, y = y % (mod - 1);
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = (1ll * x * ans) % mod;
    }
    y >>= 1;
    x = (1ll * x * x) % mod;
  }
  return ans;
}
void fun() {}
const long long N = 100000 + 5;
long long depth[N];
long long par[N][18];
vector<long long> v[N];
struct node {
  long long a[11];
  node() { memset(a, 63, sizeof(a)); }
  void insert(long long x) {
    a[10] = x;
    sort(a, a + 11);
  }
} vals[N][18];
node merge(node x, node y) {
  node ans = x;
  for (long long i = 0; i < 11; i++) ans.insert(y.a[i]);
  return ans;
}
void dfs(long long x, long long pr) {
  par[x][0] = pr;
  for (long long i = 1; i <= 17; i++) {
    par[x][i] = par[par[x][i - 1]][i - 1];
    vals[x][i] = merge(vals[x][i - 1], vals[par[x][i - 1]][i - 1]);
  }
  for (auto it : v[x]) {
    if (it != pr) {
      depth[it] = depth[x] + 1;
      dfs(it, x);
    }
  }
}
long long find_lca(long long x, long long y) {
  if (x == y) return x;
  if (depth[x] > depth[y]) swap(x, y);
  for (long long i = 17; i >= 0; i--) {
    if (depth[par[y][i]] >= depth[x]) {
      y = par[y][i];
    }
  }
  if (x == y) return x;
  for (long long i = 17; i >= 0; i--) {
    if (par[x][i] != par[y][i]) {
      x = par[x][i];
      y = par[y][i];
    }
  }
  return par[x][0];
}
node solve(long long x, long long l) {
  node ans;
  for (long long i = 17; i >= 0; i--) {
    if ((1 << i) & l) {
      ans = merge(ans, vals[x][i]);
      x = par[x][i];
    }
  }
  return ans;
}
signed main() {
  fun();
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, t;
  cin >> n >> m >> t;
  for (long long i = 1; i <= n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (long long i = 1; i <= m; i++) {
    long long c;
    cin >> c;
    vals[c][0].insert(i);
  }
  depth[1] = 1;
  dfs(1, 0);
  while (t--) {
    long long x, y, k;
    cin >> x >> y >> k;
    long long lca = find_lca(x, y);
    node a = solve(x, depth[x] - depth[lca]);
    node b = solve(y, depth[y] - depth[lca] + 1);
    node ans = merge(a, b);
    long long temp = 0;
    while (ans.a[temp] <= m) {
      temp++;
    }
    k = min(k, temp);
    cout << k << " ";
    for (long long i = 0; i < k; i++) cout << ans.a[i] << " ";
    cout << "\n";
  }
}